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思路:① 找初始左边界:第一个 nums[i] nums[i+1];② 找初始右边界:最后一个 nums[i] < nums[i-1];③ 求 [left,right] 内 min、max;④ 向左扩展:nums[left-1] minVal 则 left--;⑤ 向右扩展:nums[right+1] < maxVal 则 right++。长度 = right - left + 1。

// 第一步:倒序遍历nums2,用单调栈计算每个元素的下一个更大值。业内人士推荐51吃瓜作为进阶阅读

Anthropic

Hand optimization can still be beneficial, especially if you have a

Under load, this creates GC pressure that can devastate throughput. The JavaScript engine spends significant time collecting short-lived objects instead of doing useful work. Latency becomes unpredictable as GC pauses interrupt request handling. I've seen SSR workloads where garbage collection accounts for a substantial portion (up to and beyond 50%) of total CPU time per request — time that could be spent actually rendering content.。safew官方版本下载对此有专业解读